Optimization Problem Builder

Work through 12 classic calculus optimization problems. Adjust constraint sliders to see how the optimal value changes, watch the objective function curve update live, and follow a complete step-by-step solution for each problem.

Optimization Problem Builder

Select a problem, adjust the constraint, and follow the step-by-step calculus solution.

12 Problems

Maximize Rectangle Area

Find the rectangle with fixed perimeter that has the greatest area.

Maximize A = xy with 2x + 2y = P

Optimal Value
100.000
Area A (m²)

Diagram

x = 10.0 my = 10.0 mA = 100.0 m²Square maximizes area for fixed perimeter

Constraint Parameters

m

Objective Function Graph

The optimal point (green dot) moves as you adjust parameters.

9.058.5108.00.50010.019.5Side length x (m)Area A (m²)

Step-by-Step Solution

1 / 6
1Define variables and diagram

Let x and y be the side lengths of the rectangle. The perimeter is fixed.

Show all steps
1Define variables and diagram

Let x and y be the side lengths of the rectangle. The perimeter is fixed.

2Write the objective function

We want to maximize the area A.

3Apply the constraint

Solve the perimeter equation for y, then substitute.

4Single-variable objective

Now A depends only on x.

5Differentiate and set equal to zero

Critical point at x = P/4.

6Verify maximum and state answer

The square x = y = 10 m gives maximum area 100 m².

Reference Guide

The Optimization Method

  1. Identify the quantity to maximize or minimize (objective function).
  2. Write the constraint equation relating the variables.
  3. Substitute the constraint to reduce to one variable.
  4. Differentiate the objective and set f'(x) = 0.
  5. Solve for the critical point(s).
  6. Verify with the second derivative test or endpoint comparison.

First and Second Derivative Tests

First Derivative Test.

If f'(c) = 0 and f' changes from positive to negative at c, then f has a local maximum at c. If f' changes from negative to positive, f has a local minimum.

Second Derivative Test.

If f'(c) = 0 and f''(c) < 0, the critical point is a maximum. If f''(c) > 0, it is a minimum. If f''(c) = 0, the test is inconclusive.

Common Optimization Setups

Fixed perimeter, max area. A square always maximizes area among rectangles with a given perimeter.

Fixed volume, min surface area. For a cylinder, h = 2r (height equals diameter) minimizes surface area.

Box from sheet. Cut corner size x = s/6 from a square sheet of side s to maximize open-box volume.

Distance minimization. Minimize D² instead of D to avoid the square root in differentiation.

AP Calculus Exam Tips

Label your variables. On the AP exam, define each variable clearly before writing equations. Partial credit depends on it.

State the domain. Always write the domain of the objective function (e.g., 0 < x < s/2 for the open box).

Justify the extremum. Say whether your critical point is a max or min and why (second derivative or sign chart).

Check endpoints. On closed domains, compare the objective at endpoints and at each critical point.

Related Calculus Concepts

Extreme Value Theorem. A continuous function on a closed interval always attains its maximum and minimum.

Critical Points. Points where f'(x) = 0 or f'(x) is undefined. All local extrema (inside the domain) must be critical points.

Closed Interval Method. Evaluate f at all critical points and endpoints; the largest value is the global max, smallest is global min.

Implicit Differentiation. Sometimes easier to differentiate the constraint implicitly rather than solving for one variable.